1,1

Most probably a(5) = 1118176194, because it is a starting point of a string of 5 zeros, but the fact that this is the least such number needs to be confirmed. Note that zeros of A046694(n) have the indices equal to the terms of arithmetic progressions of the type k*p, where primes p belong to A134671 = {1381,5527,8291,12437,22111,29021,30403,34549,37313,...} = Primes of the form 2m*691 - 1. Thus: a(1) = 1381 = 2*691 - 1, a(2) = 16581 = 3*5527 = 3*(8*691 - 1), a(3) = 290217 = 3*96739 = 3*(140*691 - 1), a(4) = 1409635 = 5*281927 = 5*(408*691 - 1), a(5) = 1118176194 = 6*186362699 = 6*(269700*691 - 1). Also, note that all listed terms have the form a(n) = k*p - 1, where prime p is a prime of the form p = 2m*691 - 1 that belong to A134671. a(1) = 2*691 - 1, a(2) = 2*8291 - 1, a(3) = 2*145109 - 1, a(4) = 4*352409 - 1, a(5) = 5*223635239 - 1.

Eric Weisstein Link to a section of The World of Mathematics. Ramanujan's Tau Function.

a(1) = 1381 because A046694(1381) = 0 is the first zero in A046694(n).

a(2) = 16581 because A046694(16581) = A046694(16582) = 0 are the first two consecutive zeros in A046694(n).

Cf. A046694 = Ramanujan tau numbers mod 691 = sum of 11-th power of divisors mod 691. Cf. A134671 = Primes of the form 2m*691 - 1. Cf. A121733 = Numbers n such that two consecutive Ramanujan tau numbers are congruent mod 691. Cf. A121734 = Ramanujan tau numbers such that A000594[n] == A000594[n+1] mod 691. Cf. A121742 = Numbers n such that three consecutive Ramanujan tau numbers are congruent mod 691. Cf. A121743 = Values of the Ramanujan tau triplets mod 691 such that three consecutive Ramanujan tau numbers are congruent mod 691.

Adjacent sequences: A134667 A134668 A134669 this_sequence A134671 A134672 A134673

Sequence in context: A031796 A020406 A134671 this_sequence A092128 A029819 A060062

hard,uned,nonn

Alexander Adamchuk (alex(AT)kolmogorov.com), Nov 05 2007